1980 USAMO Problems/Problem 4

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The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.


Let $ABCD$ be the tetrahedron, and let $P$ and $Q$ be the points at which the insphere touches faces $ABC$ and $ABD$ respectively (and therefore the centroids of those faces). Looking at the plane containing $P$, $A$, and $Q$, we see that the intersection of the sphere and the plane is a circle, and that $AP$ and $AQ$ are both tangent to said circle. $[AP]$ and $[AQ]$ are tangents to the circle from the same point and thus have the same length. The same goes for $[BP]$ and $[BQ]$. Thus, triangles $ABP$ and $ABQ$ are congruent, and $\mathbf{\angle PAB = \angle QAB}$.

Let the intersection of $AP$ and $BC$ be $M$, and let the intersection of $AQ$ and $BD$ be $N$. Then $AM$ and $AN$ are medians of $ABC$ and $ABD$, and thus $|AM| = \frac{3}{2}|AP| = \frac{3}{2}|AQ| = |AN|$. We already know from the previous congruence that $\angle{MAB} = \angle{PAB} = \angle{QAB} = \angle{NAB}$, and $|AB|$ is equal to itself. Thus, $MAB$ and $NAB$ are also congruent to each other. Finally, $|BC| = 2|BM| = 2|BN| = |BD|$ (Because $M$ and $N$ are midpoints of $BC$ and $BD$ respectively), and from the congruence of $MAB$ and $NAB$ we have $\angle{CBA} = \angle{MBA} = \angle{NBA} = \angle{DBA}$, and again $|AB|$ is equal to itself. Thus $ABC$ and $ABD$ are congruent, thus $\mathbf{|AC| = |AD|}$ and $\mathbf{|BC| = |BD|}$.

By applying the same logic to faces $BCD$ and $ACD$ we get $|AC| = |BC|$ and $|AD| = |BD|$. Finally, applying the same logic to faces $ACB$ and $ACD$ we get $|AB| = |AD|$ and $|CB| = |CD|$. Putting all these equalities together, we get that all edges of the tetrahedron are equal, and thus the tetrahedron is regular. $\blacksquare$

See Also

1980 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions

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